I have data that measures light intensity with a number of different
techniques. One of the measurements (a direct measurement and "true"
measurement of light intensity) involves lots of time, labor, and expense.
The other techniques are more practical in the sense of time and labor, but
are indirect measurements (based on canopy structure (density, location of
holes in the canopy, etc). My goal is to determine if the indirect
measurements are valid estimates of the direct measurements. However, I would
also like to predict light intensity based on the indirect methods.
I can see two methods of analysis for this situation: correlation and
regression. It seems that correlation would be the best option to validate
the measurements to each other. If the measurements are correlated then the
use of regression analysis would yield a prediction equation.
For the correlation analysis, I can use Pearson or Spearman analysis. To use
Pearson, the variables should be normally distributed. However, I have read
that the distribution for correlation should be bivariate normally
distributed. I understand how to test for normality with a univariate normal
distribution but have no idea how to test for bivariate normal distribution. I
am using the SAS program to do my analysis. Does anyone know how to test for
bivariate normal distribution?
If the variables are bivariate normally distributed then I use Pearson, but if
they are not normally distributed I use Spearman. Is this correct?
The regression analysis is also somewhat confusing. Regression analysis is
based on the fact that the Y (dependent variable) is random and the X
(independent variable) is fixed with no error. For my case, both X and Y are
random and have some measurement error. Is it correct to use simple linear
regression for this analysis or is there another type of analysis to obtain
predictions?
I apologize for such a long post, but I have been struggling with this
analysis for sometime and the more information I obtain from Statistics books,
the more confused I get.
Thanks in advance, Mike
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